How to prove the integral $\int_{0}^{1} \frac{(1+x^{2})\ln(1+x)}{x^{4} -x^{2} +1}\,dx\ = \frac{\pi \ln(2+\sqrt{3} )}{6} $

Question

I am to prove that $$\int_{0}^{1} \frac{(1+x^{2})\ln(1+x)}{x^{4} -x^{2} +1}\,dx\ = \frac{\pi \ln(2+\sqrt{3}) }{6}$$ I tried to split it, into $$\frac{1}{2} (\int_{0}^{1} \frac{\ln(1+x)}{x^{2}-\sqrt{3}x+1}\,dx\ +\int_{0}^{1} \frac{\ln(1+x)}{x^{2}+\sqrt{3}x+1}\,dx) $$ but I can't figure out how to solve it further. Any help would be appreciated!